This is not true. Consider sequence $(x_n,y_n)=(n^{-2},n^{-1})$ then you get
$$
\lim\limits_{n\to\infty}(x_n,y_n)=0\\
\lim\limits_{n\to\infty}\frac{x_n y_n^2}{x_n^{2}+y_n^{4}}=1
$$
If you consider another sequence $(x_n,y_n)=(n^{-1},n^{-1})$ then you get
$$
\lim\limits_{n\to\infty}(x_n,y_n)=0\\
\lim\limits_{n\to\infty}\frac{x_n y_n^2}{x_n^{2}+y_n^{4}}=+\infty
$$
So we conclude that limit
$$
\lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}
$$
even doesn't exist, not to mention it is equal to $0$.

Let $x=y^2$ therefore we have
$$\lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}=\lim\limits_{y\to0}\frac{y^4}{2y^4}=\frac{1}{2} \ \ \ (1)$$ and if we consider $x=y$ then
$$\lim\limits_{(x,y)\to(0,0)}\frac{xy^2}{x^2+y^4}=0 \ \ \ \ (2)$$ so that from (1) and (2) we see that limit don't exits